Question

The mean and the standard deviation of a data of $8$ items are $25$ and $5$ respectively. If two items $15$ and $25$ are added to this data, then the variance of the new data is

• A. 29
• B. 24
• C. 26
• D. $\sqrt{29}$

Answer 1

Answer: (A)

We have, $n=8,\bar{x}=25$ and $\sigma =5$
$\therefore \bar{x}=\frac{\Sigma x_i}{n}$
$\Rightarrow \Sigma x_i=n\bar{x}=8\times 25=200$
$\Rightarrow$ In corrected $\Sigma x_i=200$
and $\sigma =5$
$\Rightarrow {\sigma }^2=25$
$\Rightarrow \frac{1}{n}\Sigma x_i^2-(\text{ mean }{)}^2=25$
$\Rightarrow \frac{\Sigma x_i^2}{8}-625=25$
$\Rightarrow \Sigma x_i^2=5200$
Corrected $\Sigma x_i^2=5200+225+625=6050$
and corrected mean $=200+15+25=240$
$\therefore$ Corrected variance $=\frac{6050}{10}-{\left(\frac{240}{10}\right)}^2$
$=605-(24{)}^2=605-576=29$